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B
3
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C
27
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D
18
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Solution
The correct option is C27 Consider 1r1,1r2,1r3 Now applying AM≥GM, we get 1r1+1r2+1r33≥1(r1.r2.r3)13 But, 1r1+1r2+1r3=1r ⇒13r≥1(r1.r2.r3)13 r1.r2.r3r3≥27